# Tag Info

4

For part 2., one can do without $\sigma$-finiteness and in place assume that $\{|f|\neq0\}$ is $\sigma$-finite. Assume $1<p<\infty$. For any $E\in\mathscr{F}$ with $E\subset\{|f|\neq0\}$ and $\mu(E)<\infty$, we will show that $\|f\mathbb{1}_E\|_p\leq M_f$. This would imply that $f\in\mathcal{L}_p$ and that $\|f\|_p\leq M_f$, because of the ...

2

It well known that there is a non-constant continuous function $g$ with $g'=0$ a.e. [ You can search for 'Cantor Function']. We may suppose $g(0)=0$ (by considering $g(x)-g(0)$). The function $g^{n}(x)e^{2x}$ satisfies the stated property for any positive integer $n$ so there are infinitely many solutions. Now let $f$ and $g$ be two solutions which are ...

2

Indeed you can do that. The cost of doing so is then you don't really know whether a function is integrable unless it is a simple function, so you need to work to provide a better characterization of functions in $L^1(X;E)$.

4

I think the reason people are interested in Bochner integrable functions is because people are interested in Banach space valued functions, rather than the properties of the Banach space of Bochner integrable functions. For example, you might want to do harmonic analysis or probability theory on Banach valued functions. Otherwise, why even define real ...

1

Here is a slightly general result that includes the OP with $p=2=q$. Theorem: If $1/p +1/q=1$, $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$ and $g\in\mathcal{L}_q(\mathbb{R}^n,\lambda_n)$, then $f*g$ is uniformly continuous. If $1<p<\infty$ then $f*g\in\mathcal{C}_0(\mathbb{R}^n)$. Here is a short proof: We use $\operatorname{supp}(f)$ to denote the ...

0

For all $f\in L^p$ we have $\|f(x+k)-f(x)\|_p\to 0$ when $k\to 0$. If you have never seen this statement before, the proof goes like this: first show it for step functions, then use the fact that they are dense in $L^p$ to generalize it. This result is used a lot in problems about $L^p$ spaces. Alright, so let $\varepsilon>0$. There is some $\delta>0$ ...

0

Hint: Use Fatou's Lemma and definition of derivative to show that $f(b)-f(a) \leq \int_a^{b} f'(t)dt$ whenever $a <b$. Now use the fact that $f'$ is integrable so given $\epsilon >0$ there exists $\delta >0$ such that $\int_E f'(t)dt <\epsilon$ whenever $m(E) <\delta$. Absolute continuity of $f$ on finite intervals now follows easily.

0

I notice that $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)^{1/p} \xrightarrow{\color{red}{a} \to +\infty}0}$ $\implies$ $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)^{1/p} \xrightarrow{x \to +\infty}0}$, so for all $\varepsilon>0$, $\exists$ $a_{\varepsilon}$ : $\forall x\geqslant a_{\varepsilon}$, $\displaystyle{ \left(\int_a^x|f(t)|^pdt \right)... 0 Notice that $$\sum^\infty_{n=1}n\mathbb{1}_{\{n-1\leq |f(x)| <n\}}\leq |f(x)|+1\tag{1}$$ and that $$|f(x)|\leq \sum^\infty_{n=1}n \mathbb{1}_{\{n-1\leq |f(x)| <n\}}\tag{2}$$ Since the measure is finite, integrability of$|f|$implies the integrability of$G(x)=\sum_{n\geq1}n\mathbb{1}_{E_n}$. Conversely, the integrability of$G$implies the ... 0 Yes, you can just split up into positive and negative parts. As long as you make sure that they are both integrable. It would probably be helpfull for you to give an example that you at present do not know how to handle. 1 Let$\mu$be a complex measure on a measurable space$(X,\mathscr{B})$and let$G\subset\mathbb{C}$be open. Suppose$f$is a complex valued function in$G\times X$such that$f(x,\cdot)$is holomorphic in$G$for each$x\in X$, that$f(\cdot,z)$is$\mathscr{B}$--measurable for each$z\in G$, and that$|f(x,z)|\leq g(x)$for all$(x,z)\in X\times G$and ... 1 Let$f(\alpha)$be the integral. By the mean value theorem, one has \begin{equation} |e^{-\alpha x} - e^{-\beta x}|\le |\alpha - \beta|\,x\, e^{-\min(\alpha, \beta)x} \end{equation} as$|\frac{sin x}{x}|\le 1$, one deduces \begin{equation} |f(\alpha) - f(\beta)|\le \int_0^{+\infty} |\alpha - \beta|\,x\, e^{-\min(\alpha, \beta)x}\, d x = \frac{|\alpha-\beta|}{... 1 You were almost there. Hint:$\int_0^1 f\, dm \ge \int_F f\,dm.$2 We know that$x \in \limsup_{n\to\infty} A_n$if and only if$x \in A_n$for infinitely many$n \in \Bbb{N}$. We claim that $$\limsup_{n\to\infty }A_n = (-1,+\infty).$$ Indeed, let$x \in (-1,+\infty)$. If$x \in [0,+\infty)$then$x \in A_n$for every even$n\in\Bbb{N}$such that$n \ge x$. If$x \in (-1,0)$then pick$n_0\in\Bbb{N}$such that$-1+\frac1{...

2

Yes it is. Here you can compute the value of the integral directly using monotone convergence: $$\int|f|=\int\lim_{n\rightarrow\infty} \sum^N_{k=10}2^{-k}\mathbb{1}_{[k,k+1})=\lim_{n\rightarrow\infty}\int\sum^n_{k=1}2^{-k}\mathbb{1}_{[k,k+1)}=\lim_n\sum^n_{k=0}2^{-k}=2<\infty$$

4

Observe that \begin{align*} \int_{\mathbb{R}}|f| &=\int_{\mathbb{R}} \left|\sum_{n=0}^{\infty} 2^{-n}\chi_{[n,n+1)} \right|\\ &= \int_{\mathbb{R}} \sum_{n=0}^{\infty}2^{-n}|\chi_{[n,n+1)}|\\ &=\sum_{n=0}^{\infty}2^{-n} \int_{\mathbb{R}}\chi_{[n,n+1)}&\because\text{Monotone Convergence Theorem}\\ &=\sum_{n=0}^{\infty}2^{-n}\\ &=\frac{...

0

Here we present a solution that adapts the proof of Riemann-Lebesgue lemma. Step 1. We first consider the case where $R = [a, b]\times[c, d]$ is a rectangle. Then by the Fubini-Tonelli theorem, \begin{align*} \int_{R} \cos(nxy)\,\mathrm{d}x\mathrm{d}y &= \int_{c}^{d} \left( \int_{a}^{b} \cos(nxy) \,\mathrm{d}x \right) \mathrm{d}y \\ &= \int_{c}^{d} (...

1

For $1<p<\infty,$ we can use Holder: $$\int_X|f_n-f|\,d\mu = \int_X|f_n-f|\cdot 1\,d\mu$$ $$\le (\int_X|f_n-f|^p\,d\mu)^{1/p}\cdot (\int_X 1^q\,d\mu)^{1/q}$$ $$= (\int_X|f_n-f|^p\,d\mu)^{1/p}\mu(X)^{1/q} \to 0.$$ An example where $\mu(X)=\infty$ and the result fails: On $[1,\infty)$ with Lebesgue measure, define $$f_n(x) = \frac{1}{nx}.$$ Then for $1&... 0 Assuming that$Q$is the set of rational numbers, you are given that$f$is equal almost everywhere to a continuous function, and is therefore measurable. Can you prove that if$g$is measurable,$f$is a function, and$\lambda(\{f \not= g\}) = 0$, then$f$is measurable too? 1 We have$f \in \mathcal{L}^2(\Omega)$because $$\int f^2 d\mu = \sum_{n = 1}^{\infty} f(n)^2 \mu(\{n\}) = \sum_{n = 1}^{\infty} \frac1n\cdot \frac1{\sqrt{n}} < \infty.$$ We have$f\not\in \mathcal{L}^1(\Omega)$because $$\int f d\mu = \sum_{n = 1}^{\infty} f(n) \mu(\{n\}) = \sum_{n = 1}^{\infty} \frac1{\sqrt{n}}\cdot \frac1{\sqrt{n}} = \sum_{n = 1}^{\... 2 You need f to be measurable. Let C_n=\{x:f(x)\ge 2^n\} and consider the indicator function I_{C_n}. Define$$g(x)=\sum_{n=0}^\infty 2^nI_{C_n}(x).$$Then$$\int_E g=\sum_{n=0}^\infty 2^nm(C_n).$$So you want to show \int_Ef<\infty iff \int_E g<\infty. If 2^N\le f(x)<2^{N+1} then g(x)=1+2+4+\cdots+2^N=2^{N+1}-1. If f(x)<1 then g(x)... 3 We can show that for every n \ge 2 there is an interval I_n of length between 1/n and 2/n s.t. \int_{I_n} f(x)dx=0 Fix n \ge 2 and consider a_k=\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx, k=0,1..,n-1; since \sum a_k=0 we either have some a_k =0 so done or there are consecutive a_ka_{k+1} <0 for some k \le n-2; wlog assume a_k >0, ... 5 As$$\int_0^\infty\frac{dx}{|x|+1}=\int_0^\infty\frac{dx}{x+1}=\infty,$$for all n, there is an a_n>n with$$\int_n^{a_n}\frac{dx}{x+1}=1.$$Let f_n(x) equal 1/(x+1) on the interval [n,a_n] and 0 elsewhere. 2 There is a stronger result: Suppose \sin(a_nx) converges pointwise on a set of positive measure. Then a_n converges to a finite limit. Proof: Let E be a set of positive and finite measure where \sin(a_{n}) converges pointwise. We first prove (a_n) must be bounded. If not, then WLOG there exist 0<a_{n_1} < a_{n_2} < \cdots \to \infty. By ... 5 I think the answer is meant to be \{\infty\}. Suppose$$ \int_{\mathbb R} \sup_{k} f_k = I < \infty. $$By the monotone convergence theorem, there exists a positive integer n such that$$ \int_{[-n,n]} \sup_{k} f_k \ge I-\tfrac13. $$Now consider f_{6n}. Since \sup_{x\in\mathbb R} f_{6n}(x) \le \frac1{6n}, we see that \int_{[-n,n]} f_{6n} \le \... 1 I'm assuming that \nu and \mu are positive measures, things are different for signed or complex measures. The positivity (non-negativity) assumption is one of two simple (to state) ways to ensure that both integrals appearing in (\ast) make sense. The other simple way is assuming integrability of h (with respect to \nu). But the non-negativity ... -1 First of all x is not in [0,r] but in [-r,r].You will have to see x in 4 cases. i)x\in [-r,-r/\sqrt 2] then y^2<r^2-x^2 ii)x\in [-r/\sqrt 2),0] then x<y<-x iii)x\in [0,r/\sqrt 2),] then -x<y<x iv)x\in [r/\sqrt 2),r] then y^2<r^2-x^2 1 Royden and Fitzpatrick's 4th ed put this as an exercise before they discuss absolute continuity, so I assume we can do this by Lebesgue's Differentiation Theorem which is proved via Vitali's Covering Lemma. Here's my take: Assume g \geq 0. Then f is increasing so by Lebesgue's theorem, we are done. In the general case, we can write f as a difference of ... 2 Theorem. If f\colon\mathbb R^n\to\mathbb R^n is a linear automorphism, and \lambda is the Lebesgue measure, then$$ \lambda(f(B)) = |\det f|\lambda(B)\tag{1} $$for all Borel sets B in \mathbb R^n. Proof. Decompose f as a (finite) sequence of elementary operations of the following kinds Row swap: f({\mathbf e}_i) \leftrightarrow f({\mathbf e}... 1 If there is no \delta corresponding to some \epsilon then \mu (|f_n| >\epsilon)\to 0 for some (f_n) \subseteq E_p But then 1=\int fd\mu \leq \epsilon +\int_{|f_n| >\epsilon}|f_n|d\mu. We can reach a contradiction using uniform integrability: the fact that \int |f_n|^{p} d\mu is bounded implies that (f_n) is uniformly integrable and hence ... 2 Note that since the mean and variance are finite, then$$ \sigma^2 + (EX)^2 = EX^2 = M < \infty $$Now, you know that EX^2 = \int x^2 dP_x, so as you did, you break down the integral$$ EX^2= \int_{X\leq \epsilon \sqrt n \sigma} x^2 dP_x + \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x $$Then, taking the limit$$ \lim_{n \to \infty } EX^2 = \int_{X\leq \... 0 I came up with following example where both inequalities are strict. Let$(\Omega,\mathcal{A},\mu)$be a probability space. Let$p=0.5$and$q=-1$so that$\frac{1}{p}+\frac{1}{q}=1$. Let$A\in \mathcal{A}$be such that$0<\mu(A)<1$. Let$f=1_A$and$g=1$. Then $$N_1(fg)=\mu(A)$$ $$N_p(f)=\mu(A)^2$$ $$N_q(g)=\mu(\Omega)^{-1}=1$$ so$N_1(fg)> N_p(f)...

2

Here is a sketch of the proof. I believe you can fill in the details to complete the proof: Assumption. $f$ is integrable on $[0,1]$ and satisfies $\int_{0}^{1}x^{2n}f(x)\,\mathrm{d}x=0$ for all $n\geq 0$. Then by noting that the space of even polynomials on $[0,1]$ is dense in $C([0,1])$ w.r.t. the supremum norm, we have $\int_{0}^{1}\varphi(x)f(x)\,\... 2 Make the substitution$x=\sqrt y$to see that$\int_0^{1}y^{n}g(y)dy=0$for all$n$where$g(y)=f(y)y^{-1/2}$. (Integrability of$g$follows from existence of$\int f(x)dx$which is given). This shows that$\int p(y)g(y)dy=0$for every polynomial$p$. Standard arguments using approximation of integrable functions by continuous functions and then by ... 1 We first observe that, by the Fatou's Lemma, $$\int_{1}^{\infty} f(x)^2 \, \mathrm{d}x \leq \liminf_{j\to\infty} \int_{1}^{\infty} f_j(x)^2 \, \mathrm{d}x \leq 1.$$ In particular, we obtain $$\sup_{j\in\mathbb{N}} \int_{1}^{\infty} (f(x) - f_j(x))^2 \, \mathrm{d}x \leq 4.$$ Now we fix$\epsilon \in (0, 1)$and use Egoroff's Theorem to find$E \subseteq [...

3

If $a_n\in(0,1)$ form a decreasing sequence with $\lim\limits_{n\to\infty}a_n=0$, then $$f_n(x)=\begin{cases}1/n,&0<x<a_n\\n,&a_n\leqslant x<1\end{cases}$$ satisfies the premises, and $$\int_0^1\frac{dx}{x+f_n(x)}\geqslant\int_0^{a_n}\frac{dx}{x+f_n(x)}=\log(1+na_n).$$ So, for a counterexample, it suffices to have $\log(1+na_n)/\log(1+n)\not\... 1 By Tonelli's Theorem$\int |f| \geq \int_0^{1}\int_x^{1}\frac 1 {y^{2}} dydx=\int_0^{1}(\frac 1 x-1) dx =\infty$. 2 Idea. The claim will follow if one can show that $$\lim_{h \to 0} \int_{-\infty}^{\infty} \left| \frac{f(x+h) - f(x)}{h}\right| \, \mathrm{d}x = \int_{-\infty}^{\infty} \lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h}\right| \, \mathrm{d}x,$$ since Lebesgue's Differentiation Theorem tells that$\lim_{h \to 0} \left| \frac{f(x+h) - f(x)}{h}\right| = \left| f'(...

5

Prove by contradiction. If the integral is bounded by some $M$ for all $x_0$ then $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dxdy \leq M <\infty$. By Fubini/Tonelli Theorem we can write this as $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dydx \leq M$ This is a contradiction because the inside integral is $\infty$ for every $x$.

1

$\forall\epsilon>0,\exists N$ s.t. $\forall n\ge N,m(\{x\in E\mid|f_n(x)-f(x)|>\epsilon\})<\epsilon$ $\Rightarrow \forall n\ge N,\rho(f_n,f)=\int_E \min\{1,|f_n-f|\}dm$ $=\int_{\{x\in E\mid|f_n(x)-f(x)|>\epsilon\}}1dm+\int_{E\setminus\{x\in E\mid|f_n(x)-f(x)|>\epsilon\}}|f_n-f|dm<\epsilon+\epsilon\cdot m(E)$ Thus, convergence in measure ...

1

In the backwards direction you don't have uniform convergence, only that $\int_E \min(1,|f_n-f|)\,dm \to 0$. Also, the conclusion $$\int_E \min(1,|f_n-f|)\,dm \to 0 \implies \int_E |f_n-f|\,dm \to 0$$ is not entirely spelled out in your question. Anyway, for $0 < \varepsilon < 1$ you can use Markov's inequality to obtain $$m(|f_n-f| > \varepsilon) =... 0 The author gives an explanation for that fact. The collection of half-open rectangles in \mathbb{R}^n, \mathcal{S}, is a semi-ring. So, setting \mu=\lambda^n,$$ \mu(E)=\mu^*(E)=\inf\!\left\{\sum_{j\ge 1} \mu(E_j): E\subset\bigcup_{j\ge 1} E_j,E_j\in \mathcal{S}\right\}. $$(See Eq. (6.1).) Now, by the properties of infima, for every \epsilon>0, ... 1 0. We introduce the following notation: For a finite Borel measure \mu on \mathbb{R}, we write$$G(x) = \mu((-\infty, x]).$$If f is a function on [a, b], then$$ \bigl[ f \bigr]_{a}^{b} = f(b) - f(a). $$If f is right-continuous and has left-limit at a, then we write the jump size of f at a by$$ f(a^-) = \lim_{x\uparrow a} f(x) \qquad\...

1

Hint: Hölder gives you $$\left\vert \int fg_n \right\vert \leq \int \vert fg_n \vert \leq \Vert f \chi_{E_n}\Vert_p \cdot \Vert g_n \Vert_q = \Vert f \chi_{E_n}\Vert_p,$$ which will allow you to conclude what you want.

1

Question 1. If $f \leq g$ on a set $A$, then $\int_A f \ dm \leq \int_A g \ dm$. Furthermore, if $g = c$, where $c$ is a constant, then $\int_A g = c \ m(A)$. Thus, if $f \leq c$ on $A$, with $c$ constant, then $\int_A f \ dm \leq c \ m(A)$. That's all there is to it. In the question that you linked, $f_n \leq \tfrac 3 2$ for all $n$ everywhere (and in ...

2

$$0\leq 1-\frac{x}{n}\leq e^{-x/n}\qquad 0<x\leq n$$ and so $0\leq \Big(1-\frac{x}{n}\Big)^n\mathbb{1}_{(0,n]}(x)\leq e^{-x}$. The function $x\mapsto\cos(x/n)$ is controlled easily since $|\cos|\leq 1$. By dominated convergence $$\lim_n\int^n_0\Big(1-\frac{x}{n}\Big)^n\cos(x/n)\,dx\xrightarrow{n\rightarrow\infty}\int^\infty_0 e^{-x}\,dx=1$$

0

You can see that $\int_{0}^n (1-\frac{x}{n})^n\cos{(\frac{x}{n})}dx = \int_{0}^{1}n(1-t)^{n}\cos{(t)}dt$ and you can integrate by parts

2

For each $n\in\mathbb{N}$ \begin{align} 1=\int f_n &=\int_{\{f_n\geq\frac12\}}f_n+\int_{\{f_n<\frac12\}}f_n\leq \frac32\lambda\Big(f_n\geq\frac12\Big) + \frac12\lambda\Big(f_n<\frac12\Big)\\ &=\lambda\Big(f_n\geq\frac12\Big)+\frac12 \end{align} Hence $$\lambda\Big(f_n\geq\frac12\Big)\geq\frac12\qquad\text{for all}\quad n\in\mathbb{N}$$ ...

1

To handle the $\limsup$ for your integrals over $B$, I suggest using Fatou's lemma. Fatou's lemma only applies to non-negative functions, and $\tfrac 3 2 - f_n(x)$ is a non-negative function. Applying Fatou to $\tfrac 3 2 - f_n(x)$ on $B$, we have  \int_B \liminf_{n \to \infty}\left(\tfrac 3 2 - f_n (x)\right) dm \leq \liminf_{n \to \infty} \int_B \left(\...

0

I think the integral is just equal to $\int...=\lambda(\{x\in [-1,1]^d\})= \frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)}$, if we assume euclidean norm. Then the indicator can be ignored, as it doesn't change the domain of x.

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